In A. Zee's QFT in a Nutshell, he defines the field for the Klein-Gordon equation as
$$ \tag{1}\varphi(\vec x,t) = \int\frac{d^Dk}{\sqrt{(2\pi)^D2\omega_k}}[a(\vec k)e^{-i(\omega_kt-\vec k\cdot\vec x)} + a^\dagger(\vec k)e^{i(\omega_kt-\vec k\cdot\vec x)}] $$
When calculating $\pi=\partial_0\varphi^\dagger$, I came to
$$ \tag{2}\varphi^\dagger(\vec x,t) = \int\frac{d^Dk}{\sqrt{(2\pi)^D2\omega_k}}[a^\dagger(\vec k)e^{i(\omega_kt-\vec k\cdot\vec x)} + a(\vec k)e^{-i(\omega_kt-\vec k\cdot\vec x)}] $$
But this would imply that $\varphi^\dagger=\varphi$. Is that correct?
(Intuitively it would make sense, because in QM we also consider self-adjoint operators.)
If it's correct, then why do we explicitly write $\pi=\partial_0\varphi^\dagger$ instead of just $\pi=\partial_0\varphi$? Why bother distinguishing $\varphi$ from $\varphi^\dagger$ at all?
In case it is not correct, then the first two equations of this answer are most likely wrong.
